Pattab

2022-07-16

How could I solve this chain rule problem?
To find: $\frac{d}{dt}\left[f\left(\mathbf{\text{c}}\left(t\right)\right)\right]{|}_{t=0}$
Where $\mathbf{\text{c}}\left(t\right)$ is such that $\frac{d}{dt}\mathbf{\text{c}}\left(t\right)=\mathbf{\text{F}}\left(c\left(t\right)\right)$, with $\mathbf{\text{F}}=\mathrm{\nabla }f$, $\mathbf{\text{c}}\left(0\right)=\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ and $f\left(x,y\right)=\mathrm{sin}\left(x+y\right)xy$
I think I would need to proceed using the chain rule but I am currently not sure. How would I go about solving this problem?

iskakanjulc

Expert

Let $g=f\circ c$, i.e. let
$g\left(t\right)=f\left(c\left(t\right)\right)=f\left({c}_{1}\left(t\right),{c}_{2}\left(t\right)\right).$
Then by the chain rule
${g}^{\prime }\left(t\right)={f}_{x}\left(c\left(t\right)\right){c}_{1}^{\prime }\left(t\right)+{f}_{y}\left(c\left(t\right)\right){c}_{2}^{\prime }\left(t\right)=\mathrm{\nabla }f\left(c\left(t\right)\right)\cdot {c}^{\prime }\left(t\right)=\mathrm{\nabla }f\left(c\left(t\right)\right)\cdot \mathrm{\nabla }f\left(c\left(t\right)\right),$
where the second equality follows from the assumption about ${c}^{\prime }\left(t\right)$
Since
${f}_{x}\left(x,y\right)=y\left[\mathrm{sin}\left(x+y\right)+x\mathrm{cos}\left(x+y\right)\right]$
and
${f}_{y}\left(x,y\right)=x\left[\mathrm{sin}\left(x+y\right)+y\mathrm{cos}\left(x+y\right)\right]$
we have
${f}_{x}\left(c\left(0\right)\right)={f}_{x}\left(-\frac{\pi }{2},\frac{\pi }{2}\right)=-\frac{{\pi }^{2}}{4}$
${f}_{y}\left(c\left(0\right)\right)={f}_{y}\left(-\frac{\pi }{2},\frac{\pi }{2}\right)=-\frac{{\pi }^{2}}{4}$
or
$\mathrm{\nabla }f\left(c\left(0\right)\right)=\left(-\frac{{\pi }^{2}}{4},-\frac{{\pi }^{2}}{4}\right).$
Thus
${g}^{\prime }\left(0\right)=\mathrm{\nabla }f\left(c\left(0\right)\right)\cdot \mathrm{\nabla }f\left(c\left(0\right)\right)=\left(-\frac{{\pi }^{2}}{4},-\frac{{\pi }^{2}}{4}\right)\cdot \left(-\frac{{\pi }^{2}}{4},-\frac{{\pi }^{2}}{4}\right)=\frac{{\pi }^{4}}{8}.$

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